What is the derivative of \sqrt[3]{(2+3x/(2-3x)}) ?

The derivative of \sqrt[3]{(2+3x/(2-3x)}) is 4/((3x+2)^{2/3)(-3x+2)^{4/3}}

What is the first derivative of \sqrt[3]{(2+3x/(2-3x)}) ?

The first derivative of \sqrt[3]{(2+3x/(2-3x)}) is 4/((3x+2)^{2/3)(-3x+2)^{4/3}}

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\frac{4}{( 3 x + 2 ) ^{\frac{2}{3}} ( - 3 x + 2 ) ^{\frac{4}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( \frac{2 + 3 x}{2 - 3 x} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{2 + 3 x}{2 - 3 x})

= \frac{1}{3 ( \frac{2 + 3 x}{2 - 3 x} ) ^{\frac{2}{3}}} \frac{d}{d x} (\frac{2 + 3 x}{2 - 3 x})

\frac{d}{d x} (\frac{2 + 3 x}{2 - 3 x}) = \frac{12}{( 2 - 3 x ) ^{2}}

= \frac{1}{3 ( \frac{2 + 3 x}{2 - 3 x} ) ^{\frac{2}{3}}} \cdot \frac{12}{( 2 - 3 x ) ^{2}}

Simplify

\frac{1}{3 ( \frac{2 + 3 x}{2 - 3 x} ) ^{\frac{2}{3}}} \cdot \frac{12}{( 2 - 3 x ) ^{2}} : \frac{4}{( 3 x + 2 ) ^{\frac{2}{3}} ( - 3 x + 2 ) ^{\frac{4}{3}}}

= \frac{4}{( 3 x + 2 ) ^{\frac{2}{3}} ( - 3 x + 2 ) ^{\frac{4}{3}}}

\frac{( e ^{- 3 s} )}{s ^{2}}

\frac{\partial}{\partial y} (3 x + x^{2} y)

x \to 1 lim ((a x^{2} - b x)^{2})

\frac{d}{d x} (2 c e^{2 x})

\frac{d y}{d x} = \frac{( 2 x ( y - 2 ) )}{x ^{2} + 1}, y (0) = 4

What is the derivative of \sqrt[3]{(2+3x/(2-3x)}) ?
The derivative of \sqrt[3]{(2+3x/(2-3x)}) is 4/((3x+2)^{2/3)(-3x+2)^{4/3}}
What is the first derivative of \sqrt[3]{(2+3x/(2-3x)}) ?
The first derivative of \sqrt[3]{(2+3x/(2-3x)}) is 4/((3x+2)^{2/3)(-3x+2)^{4/3}}